Question 102
What is the simplified value of $$(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)$$?
Solution
Given $$(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)$$
Multiply and divide by (3 - 1)
$$\frac{(3 - 1)(3 + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)}{3 - 1}$$
$$\frac{(3^{2} -1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)}{2}$$
$$\frac{(3^{8} - 1)(3^{8} + 1)(3^{16} + 1)}{2}$$
$$\frac{(3^{16} - 1)(3^{16} + 1)}{2}$$
$$\frac{3^{32} - 1}{2}$$
Hence, option A is the correct answer.
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